Capitulo9

Thelnward
Approach Choice
of Overall Transfer
Functions
9 .1 INTRODUCTION
In the design of control systems using the root-locus method or the frequency-domain
method, we first choose a configuration and a compensator with open parameters.
We then search for parameters such that the resulting overall system will meet design
specifications. This approach is essentially a trial-and-error method; therefore,
we usually choose the simplest possible feedback configuration (namely, a unity-feedback
configuration) and start from the simplest possible compensator-namely,
a gain (a compensator of degree O). If the design objective cannot be met by searching
the gain, we then choose a different configuration or a compensator of degree 1
(phase-lead or phase-lag network) and repeat the search. This approach starts from
interna! compensators and then designs an overall system to meet design specifica-tions;
therefore, it may be called the outward approach.
In this and the following chapters we shall introduce a different approac h, called
the inward approach. In this approach, we first search for an overall transfer function
to meet design specifications, and then choose a configuration and compute the
required compensators. Choice of overall transfer functions will be discussed in this
chapter. The implementation problem-namely, choosing a configuration and com-puting
the required compensators -will be discussed in the next chapter.
Considera plant with proper transfer function G(s) = N(s)/D(s) as shown in
Figure 9. l. In the inward approach, the first step is to choose an overall transfer
function G0(s) from the reference input r to the plant output y to meet a set of
339

340 CHAPTER 9 THE INWARD APPROACH -CHOICE OF OVERALL TRANSFER FUNCTIONS
r-
1
r 1
-i
1
1 Ga(s)
L - - -
--------,
~
1
1
- _J
Figure 9.1 Design of control systems.
specific at ion s. We claim that
G0(s) = 1
...
is the best possible system we can design. Indeed if G0(s) = 1, then y(t) = r(t), for t
;:::: O. Thus the position and velocity errors are zero; the rise time, settling time, and
overshoot are ali zero. Thus no other G0(s) can perform better than G0(s) = l. Note
that although r(t) = y(t), the power levels at the reference input and plant output
are different. The reference signa! may be provided by tuming a knob by hand; the
plant output y(t) may be the angular position of ari antenna with weight over severa}
tons.
Although G)s) = 1 is the best system, we may not be able to implement it in
practice. Recall from Chapter 6 that practica} constraints, such as proper compen-sators,
well-posedness, and total stability, do exist in the design of control systems.
These constraints impose sorne limitations in choosing G0(s). We first discuss this
problem.
9.2 IMPLEM EN T A B LE TRAN SF ER FUNCTIONS
Consider a plant with transfe r functio n
G(s)
N(s)
D(s)
where N(s) and D(s) are two polynomials and are assumed to have no common
factors. We assume n = deg D(s) ;:::: deg N(s), that is, G(s) is proper and has degree
n. An overall transfer function G0(s) is said to be implementable if there exists a
configuration such that the transfer function from the reference input r to the plant
output y in Figure 9.1 equals G0(s) and the design meets the following four
constraints:
l. Ali compensators used have proper rational transfer functions.
2. The resulting system is well-posed.
3. The resulting system is totally stable.
4. There is no plant leakage in the sense that ali forward paths from r to y pass
through the plant.
The first constraint is needed, as discussed in Section 5.4, for building compen-sators
using operational amplifier circuits. If a compensator has an improper transfer
function, then it cannot be easily built in practice. The second and third constraints

342 CHAPTER 9 THE INWARD APPROACH - - -CHOICE OF OVERALL TRANSFER FUNCTIONS
9.2 IMPLEMENTABLE TRANSFER FUNCTIONS 341
are needed, as discussed in Chapter 6, to avoid amplification of high-frequency noise
and to avoid unstable pole-zero cancellations. The fourth constraint implies that all
power must pass through the plant and that no compensator be introduced in parallel
with the plant. This constraint appears to be reasonable and seems to be met by every
configuration in the literature. This constraint is called "no plant leak:age" by
Horowitz [35].
If an overall transfer function G0(s) is not implementable, then no matter what
configuration is used to implement it, the design will violate at least one of the
preceding four constraints. Therefore, in the inward approach, the G0(s) we choose
must be implementable.
The question then is how to tell whether or nota GJs) is implementable. lt tums
out that the answer is very simple.
THEOREM 9.1
Considera plant with proper transfer function G{s) = N(s)/D(s). Then G0(s) is
implementable if and only if G0(s) and
are proper and stable. •
G0(s)
T(s) := --
G(s)
We discuss first the necessity of the theorem. Consider, for example, the con-figuration
shown in Figure 9.2. Noise, which may enter into the intput and output
terrninals of each block, is not shown. If the closed-loop transfer function from r to
y is G0(s) and if there is no plant leak:age,then the closed-loop transfer function from
r to u is T(s). Well-posedness requires every closed -loo p transfer function to be
proper, thus T(s) and G0(s) must be proper. Total stability requires every closed-loop
transfer function to be stable, thus GJs) and T(s) must be stable. This establishes
the necessity of the theorem. The sufficiency of the theorem will be established
constructively in the next chapter.
y
G(s)
Figure 9.2 Feedback system without plant leakage.

We discuss now the implication of Theorem 9.1. Let us write
G(s) = N(s)
D(s)
We assume that the numerator and denominator of each transfer function have no
common factors. The equality G0(s) = G(s)T(s) or
implies
N0(s) = N(s) . N/s)
D0(s) D(s) D1(s)
deg D0(s) - deg N0(s) = deg D(s) - deg N(s) + (deg D1(s) - deg N1(s))
Thus if T(s) is proper, that is, deg D1(s) ;:::: deg N/s), then we have
deg D0(s) - deg N0(s) 2:: deg D(s) - deg N(s) (9.1)
•
Conversely, if (9.l) holds, then deg Dr(s) 2:: deg Nt(s), and T(s) is proper.
Stability of G0(s) and T(s) requires both D0(s) and D,(s) to be Hurwitz. From
T(s) = N/s) = G0(s) = N0(s) . D(s)
D/s) G(s) D0(s) N(s)
we see that if N(s) has closed right-half-plane (RHP) roots, and if these roots are not
canceled by N0(s), then D/s) cannot be Hurwitz. Therefore, in order for T(s) to be
stable, all the closed RHP roots of N(s) must be contained in N0(s). This establishes
the following corollary.
COROLLARY 9. 1
Considera plant with proper transfer function G(s) = N(s)/D(s). Then Gjs) =
N0(s)/D0(s) is implementable if and only if
(a) deg D0(s) - deg N0(s) 2:: deg D(s) - deg N(s) (pole-zero excess inequality).
(b) All closed RHP zeros of N(s) are retained in Njs) (retainment of non-minimum-
phase zeros).
(e) D0(s) is Hurwitz. •
As was defined in Section 8.3.1, zeros in the closed RHP are called non-mini-mum-
phase zeros. Zeros in the open left half plane are called minimum-phase zeros.
Poles in the closed RHP are called unstable potes. We see that the non-minimum-phase
zeros of G(s) impose constraints on implementable G0(s) but the unstable
poles of G(s) do not. This can be easily explained from the unity-feedback config-uration
shown in Figure 9.3. Let
N(s)
G(s) = -
D(s)
be respectively the plant transfer function and compensator transfer function. Let

L----' C(s) H----..G(>) 1 1
Figure 9.3 Unity-feedback configuration.
G0(s) = N0(s)/D0(s) be the overall transfer function from the reference input r to
the plant output y. Then we have
N0(s) C(s)G(s)
G (s) = -- = -----
º D0(s) 1 + C(s)G(s)
D(s)Dc(s ) + N(s)N/s)
(9.2)
We see that N(s) appears directly as a factor of N0(s). If a root of N(s) does not
appear in NJs), the only way to achieve this is to introd u ce the same root in D(s )Dc(s)
+ N(s)Nc(s) to cancel it. This cancellation is an unstable pole-zero cancellation if
the root of N(s) is in the closed right half s-plane. In this case, the system cannot be
totally stable and the cancellation is not permitted. Therefore all non-minimum-phase
zeros of G(s) must appear in N0(s). The poles of G(s) or the roots of D(s) are shifted
to D(s)D/s) + N(s)Nc(s) by feedback, and it is immaterial whether D(s) is Hurwitz
or not. Therefore, unstable poles of G(s) do not impose any constraint on G0(s), but
non-minimum-phasezeros of G(s) do. Although the preceding assertion is developed
for the unity-feedback system shown in Figure 9.3, ít is generally true that, in any
feedback configuration without plant leakage, feedback will shift the poles of the
plant transfer function to new locations but will not affect its zeros. Therefore the
non-mínimum-phase zeros of G(s) impose constraints on GJs) but the unstable poles
of G(s) do not.
Example9.2. 1
Consíder
Then we have
G(s)
(s + 2)(s - 1)
s(s2 - 2s + 2)
G0(s) = 1 Not implementable, because it violates (a) and (b) in Corollary 9.1.
s + 2
G0(s) - (s +
3)(s + l) Not implementable , meets (a) and (e) but violates (b).
s - 1
Gº(s) - -s(_-+s_2_) Not implementable , meets (a) and (b), violates (e).
G(s)- ----- s - 1
Implementable .
o (s + 3)(s + 1)

344 CHAPTER 9 THE INWARD APPROA CH -CHOCE OF OVERALL TRANSFER 9.2 IMPLEMENTABLE F TURNACNTSOFENRS FUNCTIONS 344
s - 1
(s + 3)(s + 1)2
(2s - 3)(s - 1)
(s + 2)3
Implementab le.
Implementable.
(2s - 3)(s - l)(s + 1)
(s + 2)5
Implementable .
Exercise 9.2. l
Given G(s) = (s - 2)/(s - 3)2, are the following implementable?
1 s - 2 s - 2 (s - 2)(s - 3)
a. ;-¡---¡-- b. ;-¡---¡-- c. (s + 1)2 d. (s + 1)3
[Answers: No, no, yes, yes.]
Exercise 9.2.2
Given G(s) = (s + l)/(s - 3)2, are the following implementable?
1 s - 2 s - 2 (s - 2)s4
a. ;-¡---¡-- b. ;-¡---¡-- c. (s + 02 d. (s + 2)6
[Answers: Yes, no, yes, yes.]
From the preceding examples, we see that if the pole-zero excess inequality is
met, then ali poles and all minimum-phase zeros of G0(s) can be arbitrarily assigned.
To be precise, all poles of G0(s) can be assigned anywhere inside the open left half
s-plane (to insure stability). Other than retaining all non-minimum-phase zeros of
G(s), all minimum-phase zeros of G0(s) can be assigned anywhere in the entire
s-plane. In the assignment, if a complex number is assigned as a zero or pole, its
complex conjugate must also be assigned. Otherwise, the coefficients of G0(s) will
be complex, and G0(s) cannot be realized in the real world. Therefore, roughly speak-ing,
if G0(s) meets the pole-zero excess inequality, its poles and zeros can be arbi-trarily
assigned.
Considera plant with transfer function G(s). The problem of designin g a system
so that its overall transfer function equals a given model with transfer function Gm(s)
is called the model-matching problem. Now if Gm(s) is not implementable, no matter
how hard we try, it is not possible to match Gm(s) without violating the four con-straints.
On the other hand, if Gm(s) is implementable, it is possible, as will be shown
in the next chapter, to match Gm(s). Therefore, the model-matching problem is the
same as our implementability problem. In conclusion, in model matching, we can

344 CHAPTER 9 THE INWARD APPROA CH -CHOCE OF OVERALL TRANSFER 9.2 IMPLEMENTABLE F TURNACNTSOFENRS FUNCTIONS 345
arbitrarily assign poles as well as minimum-phase zeros so long as they meet the
pole-zero exces s inequality.
To conclude this section, we mention that if G0 is implementable, it does not
mean that it can be implemented using any configuration. For example, G0(s) =
l/(s + 1)2 is implementable for the plant G(s) = 1/ s(s - 1). This G0(s), however,
cannot be implemented in the unity-feedback configuration shown in Figure 9.3; it
can be implemented using sorne other configurations, as will be discussed in the
next chapter. In conclusion, for any G(s) and any implementable G0(s), there exists
at least one configuration in which Gs(s) can be implemented under the preceding
four constraints.
9.2.1 AsymptoticTracking and PermissiblePole-Zero
Cancellation Region
A control system with overall transfer function
/30 + f31s + /32s 2 + . . . + /3msm
G0(s) = 2
ªo + ll!¡S + 0!2S + · · · + ansn
(9.3)
with an > O and n ::::".: m, is said to achieve asymptotic tracking if the plant output
y(t) tracks eventually the reference input r(t) without an error, that is,
lim ly(t) - r(t)I = O
(---">00
Clearly if G0(s) is not stable, it cannot track any reference signal. Therefore, we
require G0(s) to be stable, which in tum requires a¡ > O for all i1• Thus, the denom-inator
of G0(s) cannot have any missing term ora term with a negative coefficient.
Now the condition for G0(s) to achieve asymptotic tracking depends on the type of
r(t) to be tracked. The more complicated r(t), the more complicated G0(s). From
Section 6.3.1, we conclude that if r(t) is a step function, the conditions for G0(s) to
achieve tracking are G0(s) stable and a0 = {30. If r(t) is a ramp function, the con-ditions
are G0(s) stable, a0 = f30; and a1 = {31. If r(t) = at', an acceleration function,
then the conditions are G0(s) stable, a0 = {30, a1 = {31 , and a2 = /32• If r(t) = O,
the only condition for y(t) to track r(t) is G0(s) stable. In this case, the output may
be excited by nonzero initial conditions, which in tum may be excited by noise or
disturbance. To bring y(t) to zero is called the regulating problem. In
conclusion, the conditions for G0(s) to achieve asymptotic tracking are simple and
can be easily met in the design.
Asymptotic tracking is a property of G0(s) as t ~ co or a steady-state property
of G0(s). It is not concemed with the manner or the speed at which y(t) approaches
r(t). This is the transient performance of G0(s). The transient performance depends
on the location of the poles and zeros of G0(s). How to choose poles and zeros to
meet the speci fic a tion on transient performance, however, is not a simple problem.
1 Also, they can ali be negative. For convenience, we consider only the posit ive case.

346 CHAPTER 9 THE INWARD APPROACH - - -CHOICE OF OVERALL TRANSFER 9.3 VARIOU SFU NDCETSIIOGNNS CRITERIA 346
In choosing an implementable overall transfer function, if a zero of G(s) is not
retained in G0(s), we must introduce a pole to cancel it in implementation. If the
zero is a non-minimum-phase zero, the pole that is introduced to cancel it is not
stable and the resulting system will not be totally stable. If the zero is minimum
phase but has a large imaginary part or is very close to the imaginary axis, then, as
was discussed in Section 6.6.2, the pole may excite a response that is very oscilla-tory
or takes a very long time to vanish. Therefore, in practice, not only the non-minimum-
phase zeros of G(s) but also those minimum-phase zeros that are close to
the imaginary axis should be retained in GJs), or the zeros of G(s) lying outside the
region C shown in Figures 6.13 or 7.4 should be retained in G0(s). How to determine
such a region, however, is not necessarily simple. See the discussion in Chapter 7.
2
Exercise 9.2.3
What types of reference signals can the following systems track without an error?
s + 5
a.
s3 +
b.
s3 +
2s2 + 8s + 5
8s + 5
2s2 + 8s + 5
2s2 + 9s + 68
c.
s3 + 2s + 9s + 68
[Answers: (a) Step functions. (b) Ramp functions. (e) None, because it is not
stable.]
9.3 VARIOUS DESIGN CRITERIA
The performance of a control system is generally specified in terms of the rise time,
settling time, overshoot, and steady-state error. Suppose we have designed two sys-tems,
one with a better transient performance but a poorer steady-state-performance,
the other with a poorer transient performance but a better steady-stage performance.
The question is: Which system should we use? This difficulty arises from the fact
that the criteria consist of more than one factor. In order to make comparisons, the
criteria may be modified as
J : = k1 X (Rise time) + k2 X (Settling time)
+ k3 X (Overshoot) + k¿ X (Steady-state error)
(9.4)
where the k¡ are weighting factors and are chosen according to the relative importance
of the rise time, settling time, and so forth. The system that has the smallest J is
called the optima/ system with respect to the criterion J. Although the criterion is

346 CHAPTER 9 THE INWARD APPROACH - - -CHOICE OF OVERALL TRANSFER 9.3 VARIOU SFU NDCETSIIOGNNS CRITERIA 347
reasonable, it is not easy to track analytically. Therefore more trackable criteria are
used in engineering.
We define
e(t) : = r(t) - y(t)
lt is the error between the reference input and the plant output at time t as shown
in Figure 9.4. Because an error exists at every t, we must consider the total error in
[O, =). One way to define the total error is
11 : = fo"' e(t)dt (9.5)
This is not a useful criterion, however, because of possible cancellations between
positive and negative errors. Thus a small 11 may not imply a small e(t) for ali t. A
better detinition of the total error is
12 : = L"' le(t)I dt (9.6)
This is called the integral of absolute error (IAE). In this case, a small 12 will imply
a small e(t). Other possible definitions are
(9.7)
Figure 9.4 Errors.

348 CHAPTER 9 THE INWARD APPROACH -CHOICE OF OVERALL TRANSFER 9.3 VARIOUS FDUENSCITGIONN CSR ITERIA 348
This arises from limited operational ranges of linear models or the physical con-straints
of devices such as the opening of valves or the rotation of rudders. Clearly,
the larger the reference input, the larger the actuating signal. For convenience, the
u(t) in (9.9) will be assumed to be excited by a unit-step reference input and the
constant Mis proportionally scaled. Now we shall check whether this const r ain t will 411
be met for all a. No matter how G0(s) is implemented, if there is no plant leakage,
the closed-loop transfer function from the reference input r to the actuating signal u
is given by
(9.11)
and
14 : = E"' tle(t)I dt (9.8)
The former is called the integral of square error (ISE) or quadratic error, and the
latter the integral of time multiplied by absolute error (ITAE). The ISE penalizes
large errors more heavily than small errors, as is shown in Figure 9.4. Because of
the unavoidable large errors at small t due to transient responses, it is reasonable not
to put too much weight on those errors. This is achieved by multiplying t with le(t)I.
Thus the ITAE puts less weight on e(t) for t small and more weight on e(t) for t
large. The total errors defined in li. 13, and 14 are all reasonable and can be used in
design.
Although these criteria are reasonable, they should not be used without consid-ering
physical constraints. To illustrate this point, we consider a plant with transfer
function G(s) = (s + 2)/ s(s + 3). Because G(s) has no non-minimum-phase zero
and has a pole-zero excess of 1, G0(s) = a/(s + a) is implementable for any positive
a. We plot in Figure 9.5(a) the responses of G0(s) dueto a unit-step reference input
for a = 1 (solid line), a = 10 (dashed line), anda = 100 (dotted line). We see that
the larger a is, the smaller 12, 13, and 14 are. In fact, as a approaches infini ty, 12, 13,
and 14 all approach zero. Therefore an optima! implementable G0(s) is a/(s + a)
with a = ce,
As discussed in Section 6. 7, the actuating signal of the plant is usually limited
by
lu(t)I :s M for all r ze O
T(s) = Gº(s)
G(s)
(9.9) ·~
(9.10)
If r is a step function, then the actuating signa! u equals
G0(s) 1 a(s + 3)
U(s) = T(s)R(s) = -- · - = -----
G(s) s (s + 2)(s + a)
This response is plotted in Figure 9.5(b) for a = 1, 10, and 100. This can be obtained
by analysis or by digital computer simulations. For this example, it happens that
lu(t)lmax = u(O) = a. For a = 100, u(O) is outside the range of the plot. We see

348 CHAPTER 9 THE INWARD APPROACH -CHOICE OF OVERALL TRANSFER 9.3 VARIOUS FDUENSCITGIONN CSR ITERIA 349
1.21-------.-----,---~--~--~--~-----.--~
; 1

1
a= 100
1
1
º[;·:-., ::=- ~ª =======<======~ o 0.5 1.5 2
0.8
a= 100
¡'' / - ----..;;:...- - - - - - - - - - - - - - - - - - - - - - - - -
i 1
i 1
a = 10
i 1
! 1
0.6 1
0.4 1
0.2
o
o 0.5
d( t)
10 i
9;
8 ~
1
7 ~
!1
6 :1
!1
5 : 1
i 1
4;1a=l0
! 1
3 ; 1
1.5 2
(a)
2.5 3 3.5 4
2
(b)
2.5 3 3.5 4
Figure 9.5 (a) Step responses. (b) Actuating signals.
that the larger a is, the larger the magnitude of the actuating signal. Therefore if a
is very large, the constraint in (9.9) will be violated.
In conclusion, in using the performance indices in (9.6) to (9.8), we must include
the constraint in (9.9). Otherwise we can make these indices as small as desired and
the system will always be saturated. Another possible constraint is to limit the band-width
of resulting overall .systems, The reason for limiting the bandwidth is to avoid
amplification of high-frequency noise. lt is believed that both constraints will lead
to comparable results. In this chapter we discuss only the constraint on actuating
signals.

350 CHAPTER 9 THE INWARD APPROACH - -CHOICE OF OVERALL TRANSFER FUNCTIONS
.
9.4 QUADRATIC PERFORMANCE INDICES
In this section we discuss the design of an overall system to minimize the quadratic
performance index
L"' [y(t) - r(t)]2 dt (9.12a)
subject to the constraint
lu(t)I :s M (9.12b)
for ali t :::::: O, and for sorne constant M. Unfortunately, no simple analytical method
is available to design such a system. Furthermore, the resulting optima! system may
not be linear and time-invariant. If we limit our design to linear time-invariant sys-tems,
then (9.12) must be replaced by the following quadratic performance index
1 = f000 [q( y(t) - r(t))2 + u2(t)] dt (9.13)
where q is a weighting factor and is required to be positive. If q is a large positive
number, more weight is placed on the error. As q approaches infinity, the contribution
of u in (9.13) becomes less significant, and at the extreme, (9.13) reduces to (9.7).
In this case, since no penalty is imposed on the actuating signal, the magnitude of
the actuating signa! may become very large or infinity; hence, the constraint in
(9. l 2b) will be violated. If q = O, then (9.13) reduc e s to
L"" u2(t)dt
and the optima! system that minimizes the criterion is the one with u = O. From
these two extreme cases, we conclude that if q in (9.13) is adequately chosen, then
the constraint in (9. l 2b) will be satisfied. Hence, although we are forced to use the
quadratic performance index in (9.13) for mathematical convenience, if q is properly
chosen, (9.13) is an acceptable substitution for (9.12).
9.4. 1 Quadratic Optimal Systems
Consid e r a plant with transfer function
G(s)
N(s)
D(s)
(9.14)
It is assumed that N(s) and D(s) have no common factors and deg N(s) :s deg D(s)
= n. The design problem is to find an overall transfer function to minimize the
quadratic performance index
J = f000 [q(y(t) - r(t))2 + u2(t)]dt (9.15)

3 51 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF9. 4O V E RQAULALO TRRAATNICSF EPRE RFFUONRCTMIOANNSC E INDICES 351
where q is a positive constant, r is the reference signa!, y is the output, and u is the
actuating signa!. Before proceeding, we first discuss the spectral factorization.
Consider the polynomial
Q(s) : = D(s)D( - s) + qN(s)N( - s) (9.16)
lt is formed from the denominator and numerator of the plant transfer function and
the weighting factor q. lt is clear that Q(s) = Q(-s). Hence, if s1 is a root of Q(s),
so is - s 1• Since ali the coefficients of Q(s) are real by assumption, if s 1 is a root of
Q(s), so is its complex conjugate sf. Consequently ali the roots of Q(s) are symmetric
with respect to the real axis, the imaginary axis, and the origin of the s-plane, as
shown in Figure 9.6. We now show that Q(s) has no root on the imaginary axis.
Consider
Q(jw) = D(jw)D( - jw) + qN(jw)N( - jw)
= ID(jw)l2 + qlN(jw)l2
(9.17)
The assumption that D(s) and N(s) have no common factors implies that there exists
no Wo such that D(JWo) = O and N(jWo) = O. Otherwise s2 + w;5 would be a
common factor of D(s) and N(s). Thus if q 7'= O, Q(jw) in (9.17) cannot be zero for
any w. Consequently, Q(s) has no root on the imaginary axis. Now we shall divide
the roots of Q(s) into two groups, those in the open left half plane and those in the
open right half plane. If all the open left-half-plane roots are denoted by D0(s), then,
because of the symmetry property, all the open right-half-plane roots can be denoted
by D0( - s). Thus, we can always factor Q(s) as
Q(s) = D(s)D( - s) + qN(s)N( -s) = D0(s)D0( -s) (9.18)
where DJs) is a Hurwitz polynomial. The factorization in (9.18) is called the spectral
factorization.
With the spectral factorization, we are ready to discuss the optima! overall trans-fer
function. The optima! overall transfer function depends on the, reference signa!
r(t). The more complicated r(t), the more complicated the optima! overall transfer
function. We discuss in the following only the case where r(t) is a step function.
lms
X d X
o
-e -b -a a b e
X -d X
Res
Figure 9.6 Distribution of the roots of Q(s) in (9.16).

352 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF9. 4O V E RQAULALO TRRAATNICSF EPRE RFFUONRCTMIOANNSC E INDICES 352
Problem Considera plant with transfer function G(s) = N(s)/D(s), as shown in
Figure 9.1, where N(s) and D(s) have no common factors and deg N(s) ::5 deg D(s)
= n. Find an implementableoverall transfer function G0(s) to minimiz e the quadratic
performance index
J = J: [q( y(t) - r(t))2 + u2(t)] dt
where q > O, and r(t) = 1 for t :::::: O, that is, r(t) is a step-reference signal.
Solution First we compute the spectral factorization:
Q(s) : = D(s)D( -s) + qN(s)N( - s) = D0(s)D0( -s)
where D0(s) is a Hurwitz polynomial. Then the optimal overall transfer function is
given by
qN(O) N(s)
(9.19)
The proof of (9.19) is beyond the scope of this text; its employment, however,
is very simple. This is illustrated by the following example.
Example 9 .4. 1
Consider a plant with transfer function
N(s)
G(s)
D(s)
1
s(s + 2)
(9.20)
Find G0(s) to minimize
J = L'° [9(y(t) - 1)2 + u2(t)]dt (9.21)
Clearly we have q = 9,
D(s) = s(s + 2) D(-s) -s(-s + 2)
and
We compute
N(s) N(-s)
Q(s) : = D(s)D( - s) + qN(s)N( - s)
s(s + 2)( -s)( -s + 2) + 9 · 1 · 1
= -s2(-s2 + 4) + 9 = s4 - 4s2 + 9
(9.22)

35 3 CHAPTER 9 THE INWARD APPROACH -CHOICE OF9 O.4V E R QALULA D TRRAANTSICFE PRE RFUFNOCRTMIOANNSC E INDICES 353
It is an even function of s. If terms with odd powers of s appear in Q(s), an error
must have been committed in the computation. Using the formula for computing the
roots of quadratic equations, we have
with
4 ± V16 - 4 · 9
s2 = -------
2
4 ± jVW
2
2 ± jVs
8 = tan-1 (~) = 48º
Thus the four roots of Q(s) are
V3e1012 = V3eJ24 º -V3eJ24º = V3e1oso0+24ºJ = V3eJ2 0 4º
V3e-J2 4º -V3e-J24 º = V3eJ(l80º-2 4 º) = V3e11s6°
as shown in Figure 9.7. The two roots in the left column are in the open right half
s-plane; the two roots in the right column are in the open left half s-plane. Using the
two left-half-plane roots, we form
Dº(s) (s + V3eJ24º)(s + V3e-J24º)
s2 + V3(ei240 + e-i240)s + 3 (9.23)
s2 + 2 · V3(cos 24º)s + 3 = s2 + 3.2s + 3
This completes the spectral factorization. Because q = 9, N(O) = 1, and D0(0)
3, the optimal system is, using (9.19),
9 . 1 1
Gº(s) = -3- . s2 + 3.2s + 3
3
s2 + 3.2s + 3
(9.24)
This G0(s) is clearly implementable. Because G0(0) = 1, the optimal system has a
zero position error. The implementation of this optimal system will be discussed in
the next chapter.
Ims
---------:0/-1<:,~--------+ Res
.................. o <,
Figure 9.7 Roots of (9.22).

354 CHAPTER 9 THE INWARD APPROACH -CHOICE OF9 O.4V E R QALULA D TRRAANTSICFE PRE RFUFNOCRTMIOANNSC E INDICES 354
..
Exercise9.4.1
Given G(s) = (s - 1)/s(s + 1), find an implementable overall transf e r function to
minimize
J = L"' [9(y(t) - 1)2 + u2(t)]dt
[Answer: G0(s) = - 3(s - l)/(s + 3)(s + l).]
(9.25)
9.4.2 Computationof SpectralFactorizations
The design of quadratic optima! systems requires the computation of spectral fac-torizations.
One way to carry out the factorization is to compute all the roots of Q(s)
and then group all the left-half-plane roots, as we did in (9.23). This method can be
easily carried out if software for solving roots of polynomials is available. For ex-ample,
if we use PC-MATLAB to carry out the spectral factorization of Q(s) in
(9.22), then the commands
q=[1o-4o9];
r== roots(q)
yield the following four roots:
r== -1.5811 +0.7071i
-1.5811-0.7071i
1.5811 +0.7071i
1.5811 - o. 7071 i
The first and second roots are in the open left half plane and will be used to form
DJs). The command
poly([r( 1) r(2)])
yields a polynomial of degree 2 with coefficients
1.0000 3.1623 3.0000
This is D0(s). Thus the use of a digital computer to carry out spectral factorizations
is very simple.
We now introduce a method of carrying out spectral factorizations without solv-ing
for roots. Consider the Q(s) in (9.22). lt is a polynomial of degree 4. In the
spectral factorization of
(9.26 )
the degrees of polynomials D0(s) and D0( - s) are the same. Therefore, the degree of
D0(s) is half of that of Q(s), or two for this example. Let

35 5 CHAPTER 9 THE INWARD APPROACH -CHOICE OF9 O.4V E R QALULA D TRRAANTSICFE PRE RFUFNOCRTMIOANNSC E INDICES 355
D0(s) = b0 + b1s + b2s2 (9.27)
where b¡ are required to be all positive.2 If any one of them is zero or negative, then
D0(s) is not Hurwitz. Clearly, we have
D0(-s) = b0 + h1(-s) + h2(-s)2 = b0 - b1s + b2s2 (9.28)
The multiplication of D0(s) and D0(-s) yields
D0(s)D0( - s) (b0 + b1s + b2s2)(b0 - b1s + b2s2)
h6 + (2b0b2 - bi}s2 + b~s4
It is an even function of s. In order to meet (9.26), we equate
b'J = 9
2b0b2 - hi = -4
and
Thus we have b0 = 3, b2 = 1 and
hi = 2b0b2 + 4 = 2 · 3 · 1 + 4 = 1 O
which implies b1 = vio. Note that we require all b¡ to be positive; therefore, we
have taken only the positive part of the square roots. Thus the spectral factorization
of (9.26) is
D0(s) = 3 + vio s + s2 = 3 + 3.2s + s2
We see that this procedure is quite simple and can be used if a digital computer and
the required software are not available. The preceding result can be stated more
generally as follows: If
and if
(9.29)
then
(9.30)
Note that before computing b1 , we must compute first b0 and b2•
(9.31)
Now we shall extend the preceding procedure to a more general case. Consider
Q(s) = ªo + a2s2 + a4s4 + a6s6
It is an even polynomial of degree 6. Let
DJs) = b0 + b,« + b2s2 + b3s3
(9.32)
(9.33)
2 Also, they can ali be negat ive. For convenience, we consider only the positive case.

356 CHAPTER 9 THE INWARD APPROA CH --CHOICE OF OVERAll TRANSFER 9.4 QUADRATIC PE RFFUONRCTMIOANNSC E INDICES 356
Then
and
D0(s)D0(-s) = bÓ + (2bob2
Equating (9.32) and (9.34) yields
and
(9.35a)
(9.35b)
(9.35c)
b~ = -a6 (9.35d)
From (9.35a) and (9.35d), we can readily compute b0 = ~ and b3 = ~. In
other words, the leading and constant coefficients of D0(s) are simply the square
roots of the magnitudes of the leading and constant coefficients of Q(s). Once b0
and b3 are computed, there are only two unknowns, b, and b2, in the two equations
in (9.35b) and (9.35c). These two equations are not linear and can be solved itera-tively
as follows. We rewrite them as
b, Y2b0 b2 - a2 (9.36a)
(9.36b)
First we choose an arbitrary b2-say, b&ºl-and use this Viºl to compute b1 as
b1) = V2b0b&º) - a2
We then use this b1 l to compute b2 as
b&1) = VG4 + 2b%3
If b&1l happens to equal b&ºl, then the chosen b&0> is the solution of (9.36). Of course,
the possibility of having b&1l = b&º) is extremely small. We then use b~1l to compute
a new b, as
and then a new b2 as
b~2) = V G4 + 2b2lb3
If b~2l is still quite different from b&n, we repeat the process. It can be shown that
the process will converge to the true solutions.3 This is an iterative method of car-rying
out the spectral factorization. In application, we may stop the iteration when
the difference between two subsequent b~) and b~+ I) is smaller than, say, 5%. This
is illustrated by an example.
3lf we compute b2 = (a2 + bf)/2b0 and b, = (b~ - a4)/2b3 iterat ively, the process will diverge.

356 CHAPTER 9 THE INWARD APPROA CH --CHOICE OF OVERAll TRANSFER 9.4 QUADRATIC PE RFFUONRCTMIOANNSC E INDICES 357
Example9.4.2
Compute the spectral factorization of
Q(s) = 25 - 41s2 + 20s4 - 4s6 (9.37)
Let
D0(s) = b¿ + b1s + b2s2 + b3s3
lts constant term and leading coefficient are simply the square roots of the corre-sponding
coefficients of Q(s):
bº = Y2s = s b3 = VR = V4 = 2
The substitution of these into (9.36) yields
b1 /1ob2 + 41
b2 /20 + 4b¡
Now we shall solve these equations iteratively. Arbitrarily, we choose b2 as b~ºl
O and compute
(0) (1) (2) (3) (4) (5)
b¡ 6.4 10.42 10.93 10.99 10.999
b2 o 6.75 7.85 7.98 7.998 7.9998
We see that they converge rapidly to the solutions b, = 11 and b2 = 8. To verify
the convergence, we now choose b2 as b~ºl = 100 and compute
(O) (1) (2) (3) (4) (5)
b¡ 32.26 12.77 11.19 11.02 11.002
b2 100 12.21 8.43 8.05 8.005 8.0006
They also converge rapidly to the solutions b, = 11 and b2 = 8.
The preceding iterative procedure can be extended to the general case. The basic
idea is the same and will not be repeated.

r¡
Exercise 9.4.2
Carry out spectral factorizations for
a. 4s4 - 9s2 + 16
b. -4s6 + 10s4 - 20s2 + 16
[Answers: 2s2 + 5s + 4, 2s3 + 6.65s2 + 8.56s + 4.]
9.4.3 Selection of Weighting Factors
In this subsection we discuss the problem of selecting a weighting factor in the
quadratic performance index to meet the constraint lu(t)I ::::: M for all t :::::: O. It is
generally true that a larger q yields a larger actuating signal and a faster response.
Conversely, a smaller q yields a smaller actuating signal and a slower response.
Therefore, by choosing q properly, the constraint on the actuating signa! can be met.
We use the exarnple in (9.20) to illustrate the procedure.
Considera plant with transfer function G(s) = 1/ s(s + 2). Designan overall
system to minimize
J = Íooo [q( y(t) - 1)2 + uz(t)]dt
lt is also required that the actuating signa! due to a unit-step reference input meet
the constraint luCt)I ::::: 3, for all t:::::: O. Arbitrarily, we choose q = 100 and compute
Q(s) = s(s + 2)(-s)(-s + 2) + 100 · 1·1 = s4 - 4s2 + 100
Its spectral factorization can be computed as, using (9.31),
D0(s) = s2 + v24s + 10 = s2 + 4.9s + 10
Thus the quadratic optima! transfer function is
Y(s) qN(O) N(s)
=--=--·--=
R(s) D0(0) D0(s)
10
s2 + 4.9s + 10
100 . 1 1
---.
10 s2 + 4.9s + 10
The unit-step response of this system is simulated and plotted in Figure 9.8(a). lts
rise time, settling time, and overshoot are 0.92 s, 1.70 s, and 2.13%, respectively.
Although the response is quite good, we must check whether or not its actuating
signa! meets the constraint. No matter what configuration is used to implement G0(s),
if there is no plant leakage, the transfer function from the reference signa! r to the
actuating signa! u is
T(s) = Gº(s)
G(s)
10 s(s + 2)
s2 + 4.9s + 10 1
lOs(s + 2)
s2 + 4.9s + 10

9.4 QUADRATIC PERFORMANCE INDICES 359
y(I) .Y( I) .Y(I)
o o o
5 5 10 5
u( t) u( t) u(t)
10 2 3
0.8
o 5 o 5 o 5
(a) (b) (e)
q = 100 0.64 9
Rise time =0.92 6.21 2.01
Set tling t ime= 1.70 10.15 2.80
Overshoot =2.13% 0% 0.09%
u(O+) =10 0.8 3
Figure 9.8 Responses of quadrat ic opt ima! systems.
The unit-step response of T(s) is simulated and also plotted in Figure 9.8(a). We see
that u(O +) = 10 and the constraint luU)I ::5 3 is violated. Because the largest mag-nitude
of u(t) occurs at t = o+, it can also be computed by using the initial -value
theorem (see Appendix A). The respon se u(t) dueto r(t) 1 is
lOs(s + 2)
U(s) = T(s)R(s) = s2 + 4.9s + 10
s
The application of the initial-value theorem yields
u(O+) = lim sU(s) = lim sT(s)R(s) = lim T(s) = 10
S-'>00
Thus the constraint lu(t)I ::::; 3 is not met and the selection of q = 100 is not
acceptable.4
Next we choose q = 0.64 and repeat the design. The optimal transfer function
is found as
0.64 X 1
0.8 s2 + v5.6 s + 0.8
0.8
s2 + 2.4s + 0.8
4It is shown by B. Seo [51] that if a plant t ransfer function is of the form (h1s + h0)/ s(s + a), with
b0 # O, then the maximum magnitude of the actuating signa! of quadrat ic opt ima! systems occurs at
t = o+ and lu<rJI :5 u(o+¡ = vq.

360 CHAPTER 9 THE INWARD APPROA CH- - -CHOICE OF OVERALL TRANSFER FUNCTIONS
Its unit-step response and the actuating signal are plotted in Figure 9.8(b). The re-sponse
is fairly slow. Because lu(t)I :s u(O+) = 0.8 is much smaller than 3, the
system can be designed to respond faster. Next we try q = 9, and compute
Q(s) = s(s + 2)( - s)( - s + 2) + 9 · 1 · 1
Its spectral factorization, using (9.31), is
D0(s) = s2 + VW s + 3 s2 + 3.2s + 3
Thus the optimal transfer function is
qN(O) N(s) 9 · 1 1
Gº(s) = D0(0) . D0(s) = -3- . s2 + 3.2s + 3
and the transfer function from r to u is
3
s2 + 3.2s + 3
19.38)
T(s) =
G0(s)
-- 3s(s + 2)
=' -----
G(s) s2 + 3.2s + 3
Their unit-step responses are plotted in Figure 9.8(c). The rise time of y(t) is 2.01
seconds, the settling time is 2.80 seconds, and the overshoot is 0.09%. We also have
iu(t)I :s u(O+) = T(oo) = 3, for all t. Thus this overall system has the fastest response
under the constraint lu(t)I :s 3.
From this example, we see that the weighting factor q is to be chosen by trial
and error. We choose an arbitrary q, say q = q0, and carry out the design. After the
completion of the design, we then simulare the resulting overall system. If the re-sponse
is slow or sluggish, we may increase q and repeat the design. In this case,
the response will become faster. However, the actuating signal may also become
larger and the plant may be saturated. Thus the choice of q is generally reached by
a compromise between the speed of response and the constraint on the actuating
signal. ·
Optimality is a fancy word because it means "the best." However, without
introducing a performance index , it is meaningless to talk about optima li ty. Even if
a performance index is introduced, if it is not properly chosen, the resul t ing system
may not be satisfactory in practice. For example, the second system in Figure 9.8 is
optimal with q = 0.64, but it is very slow. Therefore, the choice of a suitable
performance index is not necessarily simple.
Exercise 9.4.3
Given a plant with transfer function G(s) = (s + 2)/ s(s - 2), find a quadratic
optimal system under the constraint that the magnitude of the actuating signal due
to a unit step reference input is less than 5.
[Answer: G0(s) = 5(s + 2)/(s2 + 7s + 10).]

361 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF OVERALL TR9A.5N S F ETHRR EFEU N CMTOIORNES EXAMPLES 361
9.5 THREE MORE EXAMPLES
In this sectíon we shall discuss three more examples. Every one of them will be
redesigned in latter sections and be compared with quadratic optima! design.
Example 9.5.1
Considera plant with transfer function [19, 34]
2
G(s) = -------
s(s2 + 0.25s + 6.25)
Design an overall system to minimize
(9.39)
J = L"' [q(y(t) - 02 + u2(t)]dt (9.40)
The weighting factor q is to be chosen so that the actuating signal u(t) dueto a unit-step
reference input meets /u(t)/ :5 10 for t 2'. O. First we choose q = 9 and compute
Q(s) = D(s)D( - s) + qN(s)N( - s)
s(s2 + 0.25s + 6.25) · ( - s)(s2 - 0.25s + 6.25) + 9 · 2 · 2 (9.41)
= - s6 - 12.4375s4 - 39.0625s2 + 36
The spectral factorization of (9.41) can be carried out iteratívely as discussed in
Sectíon 9.4.2 or by solving its roots. As a review, we use both methods in this
example. We first use the former method. Let
D0(s) = b0 + b1s + b2s2 + b3s3
Its constant term and leading coefficient are simply the square roots of the corre-sponding
coefficients of Q(s):
b0 = V36 = 6 b3 = vFi1 = vI = 1
The substitution of these into (9.36) yields
b¡ v' 12b2 + 39.0625
b2 v'2b1 - 12.4375
Now we shall solve these equations iteratively. Arbitrarily, we choose b2 as b~ºl =
O and compute
b, 6.25 6.49 6.91 7.30 7.53 7.65 7.70 7.73 7.75 7.75
b2 o
0.25
0.73
1.18 1.47
1.62
1.69
1.72 1.74
1.74
1.75

We see that they converge to the solution b, = 7.75 and b2 = 1.75. Thus we have
Q(s) = D0(s)D0( - s) with
D0(s) = s3 + l.75s2 + 7.75s + 6
Thus the optima! overall transfer function that minimizes (9.40) with q = 9 is
Gº(s) =
qN(O) . N(s) = 9 · 2
_1 2 2 _
D0(0) D0(s) 6 s + 1.75s + 7.75s + 6
6
s3 + 1.75s2 + 7.75s + 6
(9.42)
For this overall transfer function, it is found by computer simulation that iu(t)I ~ 3,
for t 2: O. Thus we may choose a larger q. We choose q = 100 and compute
Q(s) = D(s)D( - s) + IOON(s)N( - s)
= -s6 - 12.4375s4 - 39.0625s2 + 400
Now we use the second method to carry out the spectral factorization. We use
PC-MATLAB to compute its roots. The command
r = roots([ -1 O -12.4375 O - 39.0625 O 400])
y(t)
/ Quadrat ic opt ima]
0.8
0.6
0.4
0.2
2 3 4 5 6 7 8
Figure 9.9 Responses of various designs of (9.39).

yields
t= - 0.9917+ 3.0249i
-0.9917 -3.0249i
0.9917+ 3.0249i
0.9917- 3.0249i
1.9737
-1.9737
The ñrst, secon d, and last roots are in the open left half plane. The comman d
poly([r(1) r(2) r(6)])
yields (1.000 3.9571 14.0480 20.0000]. Thus we have DaCs) = s3 + 3.957s2
+ 14.048s + 20 and the quadratic optima! overall transfer function is
20
G (s) - -=------o:--------
º - s3 + 3.957s2 + l4.048 s + 20
(9.43)
For this transfer function, the maximum amplitude of the actuating signa! due to a
unit-step reference input is 10. Thus we cannot choose a larger q. The unit-step
response of G0(s) in (9.43) is plotted in Figure 9.9 with the solid line. The response
appears to be quite satisfactory.
Exar,nple 9.5.2
s + 3
G(s) = s(s - 1) (9.44)
Find an overall transfer function to minimize the quadratic performance index
J = f000 [lOO(y(t) - 1)2 + u2(t)]dt (9.45)
where the weighting factor has been chosen as 100. We first compute
Q(s) = D(s)D( - s) + qN(s)N( - s)
= s(s - 1)(-s)(-s - 1) + lOO(s + 3)(-s + 3)
= s4 - 101s2 + 900
= (s + 9.5459)(s - 9.5459)(s + 3.1427)(s - 3.1427)
where we have used PC-MATLAB to compute the roots of Q(s). Thus we have
Q(s) = D0(s)D0( - s) with

D0(s) = (s + 9.5459)(s + 3.1427) = s2 + 12.7s + 30
and the quadratic optimal system is given by
qN(O) N(s) lO(s + 3)
Gº(s) = D0(0) . D0(s) = s2 + 12.7s + 30
o
(9.46)
Its response dueto a unit-step reference input is shown in Figure 9.lO(a) with the
solid line. The actuating signal dueto a unit-step reference input is shown in Figure
9.lO(b) with the solid line; it has the property lu(t)I ~ 10 for t;:::: O.
y(t)
1.21------------- ~
0.8
u(t)
0.6
Computer simulation
0.4
o
0.2 -2
O"---~-~--------...i....r-4'---~-~---------"-
0 0.1 0.2 0.3 0.4 0.5 0.6 0.1 0.2 0.3 0.4 0.5 0.6
(a)
(b)
Example 9.5.3
s - 1
G(s) = ---
s(s - 2)
(9.47)
lt has a non-minimum-phase zero. To find the optima! system to minimize the quad-ratic
performance index in (9.45), we compute
Q(s) s(s - 2)(-s)(-s - 2) + lOO(s - 1)(-s - 1) = s4 - 104s2 + 100
(s + 10.1503)(s - 10.1503)(s + 0.9852)(s - 0.9852)
Thus we have D0(s) = s2 + 1 l.14s + 10 and
- lO(s - 1)
G (s) - ------
º - s2 + 1 l.14s + 10
(9.48)

Its unit-step response is shown in Figure 9.11 with the solid line. By computer
simulation we also find luU)I :S 10 for t 2 ::: O if the reference input is a unit-step
function.
y(t)
~ ~T~E optima!
/ / : ..:----··· ··· -- ---··--------- -=-=-: ...:-:: ...::-; ..;-:=...;; -;...: -=-=-~-~-------j
0.5
o
-0.5
-1
-1.5 o
/ /
I / Quadrat ic opt ima!
/ /
1/
1/ Computer simulation
¡/
¡
2 3 4 5 6
9.5.1 Symmetric Root Loci5
We discuss in this subsection the poles of G0(s) as a function of q by using the root-locus
method. Consider the polynomial
D(s)D( - s) + qN(s)N( - s) (9.49)
The roots of (9.49) are the zeros of the rational function
1 + qG(s)G( - s)
or the solution of the equation
-1
q
G(s)G( - s) = N(s)N( - s)
D(s)D(-s)
(9.50)
These equations are similar to (7.11) through (7.13), thus the root-locus method can
be directly applied. The root loci of (9.50) for G(s) = 1/ s(s + 2) are plotted in
5This sect ion may be skipped without loss of continuity.

366 CHAPTER 9 THE INWARD APPROACH-CHOJCE OF OVERALL TRANSFER FUNCTIONS
9.6 ITAE OPTIMAL SYSTEMS (34) 366
/ ' - "n
x
/
Ims
' /
Figure 9.12 Root loci of (9.50).
Figure 9.12. The roots for q = 0.64, 4, 9, and 100 are indicated as shown. We see
that the root loci are symmetric with respect to the imaginary axis as well as the real
axis. Furthermore the root loci will not cross the imaginary axis for q > O. Although
the root loci reveal the migration of the poles of the quadratic optimal system, they
do not tell us how to pick a specific set of poles to meet the constraint on the actuating
signal.
We discuss now the poles of G0(s) as q--'> ca, It is assumed that G(s) has n poles
and m zeros and has no non-minimum-phase zeros. Then, as q --'> oo, 2m root
loci of G(s)G( - s) will approach the 2m roots of N(s)N( - s) and the remaining
'(2n - 2m) root loci will approach the (2n - 2m) asymptotes with angles
(2k + 1)'1T
2n 2m
k = O, 1, 2, ... , 2n - 2m - 1
lms Ims
--- ;,
X - )(,
/
/ / 3
/ / 1
/
1
X
/
' /
"X- - - X,.
n-m=3 n-m=4
Figure 9.13 Distribution of optima! potes as q -4 oo,

367 CHAPTER 9 THE INWARD APPROACH-CHOJCE OF OVERALL TRANSFER FUNCTIONS
9.6 ITAE OPTIMAL SYSTEMS (34) 367
(see Section 7.4, in particular, (7.27)). Thus as q approaches infinity, m poles of
G0(s) will cancel m zeros of G(s) and the remaining (n - m) poles of G0(s) will
distribute as shown in Figure 9.13, where we have assumed that the centroid defined
in (7.27a) is at the origin. The pole pattem is identical to that of the Butterworth
filter [13].
9 .6 ITAE OPTIMA L SYSTEMS [33]
In this section we discuss the design of control systems to minimize the integral of
time multiplied by absolute error (ITAE) in '(9.8). For the quadratic overall system
G(s) - -----
1
º - s2 + 2(s + 1
the ITAE, the integral of absolute error (IAE) in (9.6), and the integral of square
error (ISE) in (9.7) as a function of the damping ratio (are plotted in Figure 9.14.
The ITAE has largest changes as (varíes, and therefore has the best selectivity. The
ITAE also yields a system with a faster response than other criteria, therefore Graham
and Lathrop [33] chose it as their design criterion. The system that has the smallest
ITAE is called the optima! system in the sense of ITAE or the ITAE optima[ system.
Consider the overall transfer function
G (s) - --------ª~º-------
º - s" + ªn- lSn-1 + ... + ll'.zS2 + ll'.¡S + ll'.o
(9.51)
This transfer function contains no zeros. Because GJO) = 1, if G0(s) is stable, then
the position error is zero, or the plant output will track asymptotically any step-reference
input. By analog computer simulation, the denominators of ITAE optimal
systems were found to assume the forms listed in Table 9.1. Their poles and unit-step
responses, for w0 = 1, are plotted in Figures 9 .15 and 9.16. We see that the
optimal poles are distributed evenly around the neighborhood of the unit circle. We
also see that the overshoots of the unit-step responses are fairly large for large n.
These systems are called the ITAE zero-position-error optima! systems.
J4(ITAE)
4
2
o~---+----+----+-------s
0.4 0.8 1.2
Figure 9.14 Comparison of various design criteria.

368 CHAPTER 9 THE INWARD APPROACH-CHOICE OF OVERALL TRANSFER 9.6 ITAE OPTIM FAULN C STYIOSNTESM S (34) 368
Table 9.1 ITAE Zero-Position-Error Optimal Systems
p -
I
o
/ -
S + Wo
s2 + l.4Wos + w6
s3 + l.75Wos2 + 2.15wis + w5
s4 + 2.1Wos3 + 3.4wis2 + 2.7w5s + wó
s5 + 2.8Wos4 + 5.0wis3 + 5.5w6s2 + 3.4wós + w6
s6 + 3.25w0s5 + 6.60wis4 + 8.60w5s3 + 7.45wós2 + 3.95w6s + wg
s7 + 4.475Wos6 + I0.42w6s5 + 15.08w5s4 + 15.54wós3 + 10.64w6s2 + 4.58wgs + wó
s8 + 5.20Wos7 + 12.80wis6 + 21.60w6s5 + 25.75"JÓs4 + 22.20w6s3 + 13.30wgs2 + 5.I5w"/ is + w~
We now discuss the optimization of
Go(s) = sn + ªn-lsn-1 + ... + 0'.2S2 + O'.¡S + ªº (9.52)
with respect to the ITAE criterion. The transfer function has one zero; their coeffi-cients,
however, are constrained so that G0(s) has zero position error and zero ve-locity
error. This system will track asymptotically any ramp-reference input. By
analog computer simulation, the optimal step responses of G0(s) in (9.52) are found
as shown in Figure 9.17. The optima} denominators of G0(s) in (9.52) are listed in
Table 9.2. The systems are called the ITAE zero-velocity-error optima[ systems.
2nd arder 3rd
+j o - +j
/
/
/ /
4th o
/
/o
- +j
5th o
- +}
I I
o o
-1 -1

Q_
'
, -l o --1--0------1 o
-1 -1

'º-,
o
-j o -j -j
' -i
o o
6th o 7th
+j o .9 +j
.o /
/ / o
/ /
I I
o o
-1 -1
o
o
8th
o
/ - +)
/ o
/
/ o
I
--<>-----<---<,: , o
-1
o

'o
o
'
-j o o -j
' .. o
-j
o
Figure 9.15 Optima! pole locations.

368 CHAPTER 9 THE INWARD APPROACH-CHOICE OF OVERALL TRANSFER 9.6 ITAE OPTIM FAULN C STYIOSNTESM S (34) 369
5 10 15
Normalized t ime
Figure 9.16 Step responses of ITAE optima! systems with zero position error.
5 6
Normalized t ime
Figure 9.17 Step responses of ITAE optima! systems with zero velocity error.

required to track a more complicated reference input, then the transient performance
will be poorer. For example, the system in Figure 9.18 tracks acceleration reference 4
inputs, but its transient response is much worse than the one for the system in Figure
9.16, which can track only step-reference inputs. Therefore a price must be paid if
,
+ + ... + + +
~ 'I
Table 9.2 ITAE Zero-Velocity-Error Optimal Systems
s2 + 3.2Wos + w6
s3 + l.75Wos2 + 3.2Sw6s + w(i
s4 + 2.41Wos3 + 4.93w6s2 + 5.14w{is + wó
s5 + 2. l 9Wos4 + 6.50w6s3 + 6.30w5s2 + 5.24wós + w6
s6 + 6.1~Wos5 + 13.42w5s4 + 17.16wÜs 3 + 14.14wós2 + 6.76wós + w8
Similarly, for the following overall transfer function
a2s2 + a1s + a0
G (s) = (9.53)
o Sn ªn-lSn-1 azS2 a¡S ao
the optimal step responses are shown in Figure 9.18 and the optimal denominators
are listed in Table 9.3. They are called the ITAE zero-acceleration-error optima!
systems. We see from Figures 9.16, 9.17, and 9.18 that the optimal step responses
for G/s) with and without zeros are quite different. It appears that if a system is
we design a more complex system .
1.5
~O)
o :
o
~
.o...).
o.
2
Cl'J
1.0
0.5
o .. _.... _._ ,
o 5 10 15
Normalized time
Figure 9.18 Step respons es of ITAE optima! systems with zero acceleration error.

9 .6 ITAE OPTIMAL SYSTEMS (34) J71
Table 9.3 ITAE Zero-Acceleration-ErrorOptimal Systems
w0
s3 + 2.97Wos2 + 4.94w6s + w6
s4 + 3.71Wos3 + 7.88w~s2 + 5.93%s + wó
s5 + 3.81Wos4 + 9.94w5s3 + 13.44w6s2 + 7.36w(ls + w6
s6 + 3.93Wos5 + 11.68w6s4 + 18.56w6s3 + 19.3wós2 + S.06w6s + w8
9.6. 1 Applications
In this subsection we discuss how to use Tables 9.1 through 9.3 to design ITAE
optimal systems. These tables were developed without considering plant trans-fer
functions. For example, for two different plant transfer functions such as
1/s(s + 2) and l/s(s - 10), the optimal transfer function G0(s) can be chosen as
s2 + 1.4Wos + w6
The actuating signals for both systems, however, will be dífferent. Therefore w0 in
both systems should be dífferent. We shall use the constraint on the actuating signal
as a criteríon in choosing %· This will be illustrated in the following examples.
Example 9.6.1
Consider the plant transfer function in (9.20) or
1
G(s) - -s(_s_+_2_)
Find a zero-position-error system to mínimize IT AE. It is also required that the
actuating signal due to a unit-step reference input satisfy the constraint
for all t.
iu(t)I :5 3
The IT AE optimal overall transfer function is chosen from Table 9 .1 as
w6
Gº(s) =
s
2 + 1.4w0s + 2
lt is implementable. Clearly the larger the w0, the faster the response. However, the
actuating signal will also be larger. Now we shall choose Wo to meet iu(t)I :5 3. The
transfer function from r to u is
U(s)
T(s) := -
R(s)
G0(s)
G(s)

372 CHAPTER 9 THE INWARD APPROACH - --CHOICE OF OVERALL TRANSFER FUNCTIONS
s w6
y(t) u(t)
+
1.41----~,----------~
'
x 1.2
'
2.5
2
1.5
0.8
ITAE zero-velocity-error
0.6
0.4
0.2
0.5
o
-0.5
,_/ /
o~---~--~~-~~-~- - - ' - - 1 -1~~-~---~--~~ -~ ,
o 4 10 o 4 10
(a) (b)
Figure 9.19 Step responses of (9.54) (with solid lines) and (9.55) (with dashed lines).
Conse qu e ntly , we have
U(s) = T(s)R(s) =
%s(s + 2) 1
s2 + 1.4Wos + w6 s
By computer simulation, we find that the largest magnitude of u(t) occurs at t
o+ .6 Thus the Iargest magnitude of u(t) can be computed by using the initial-value
theorem as
w6s(s + 2)
= u(O+) = lim sU(s ) = lim -2---=------
s-"oo s-"oo + 1.4w0s +
In arder to meet the constraint lu(t)I 5 3, we set w6
system is
3. Thus the ITAE optima!
3 3
(9.54)
Gº(s) = s 2 + 1.4 X
• r,:3s 3
V .i s2 + 2.4s + 3
This differs from the quadratic optima! system in (9.38) only in one coefficient.
Because they minimize different criteria, there is no reason that they have the same
overall transfer function. The unit-step responses of G0(s) and T(s) are shown in
Figure 9.19 with solid lines. They appear to be satisfactory.
Exercise 9.6.1
Consider a plant with transfer function 2/ s2• Find an optima! system to minimize
the ITAE criterion under the constraint lu<t)I 5 3.
[Answer: 6/(s2 + 3.4s + 6).]
6If
the largest magnitude of u(t) does not occur at t = O, then its analytical computation will be com-plicated.
It is easier to find it by computer simulations.

373 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF OVERAL9L. 6 T R AITNAES F EORP TFIUMNACLT ÍSOYNSSTE MS (34) 373
+ +
(9.55)
Example 9.6.2
Consider the problem in Example 9.6.1 with the additional requirement that the
velocity error be zero. A possible overall transfer function is, from Table 9.2,
3.2Wos + w6
Gº(s) = 2 2
s 3.2w0s Wo
However, this is not implementable because it violates the pole-zero excess inequal-ity.
Now we choose from Table 9.2 the transfer function of degree 3:
3.25w6s + w6
Gº(s) =
S
3
+ 1.75WoS
2
+ 2
3.25WoS
3
+ Wo
This is implementable and has zero velocity error. Now we choose w0 so that the
actuating signal due to a unit-step reference input meets iu(t)I :S 3. The transfer
function from r to u is
T(s) =
G0(s)
--
(3.25w6s + w6)s(s + 2)
= -----"----"------
G(s) s3 + 1.75Wos2 + 3.25w6s + w6
lts unit-step response is shown in Figure 9.19(b) with the dashed line. We see that
the largest magnitude of u(t) does not occur at t = o+. Therefore, the procedure in
Example 9.6.1 cannot be used to choose w0 for this problem. By computer simula-tion,
we find that if áJo = 0.928, then /u(t)/ .5 3. Por this áJo, C:,(s) becomes
2. 799s + O. 799
G (s) - --=----..,....-------
º - s3 + 1.624s2 + 2.799s + 0.799
This is the IT AE zero-velocity-error optimal system. Its unit-step response is plotted
in Figure 9.19(a) with the dashed line. lt is much more oscillatory than that of the
ITAE zero-position-error optimal system. The corresponding actuating signal is plot-ted
in Figure 9.19(b).
Example 9.6.3
Consider the plant transfer function in (9.39), that is,
2
G(s) = -s(-s2-+-0 .-2-5s _+ _6.-2-5)
Find an IT AE zero-position-error optimal system. It is also required that the actuating
signal u(t) due to a unit-step reference input meet the constraint iu(t)I :S 10, for
t ::=::::O. We choose from Table 9.1

374 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF OVERAL9L. 6 T R AITNAES F EORP TFIUMNACLT ÍSOYNSSTE MS (34) 373
By computer simulation, we find that if Wo = 2.7144, then luCt)I ::; u(O)
ali t 2:: O. Thus the ITAE optimal system is
0
10 for
G (s) - ----------
20
(9.56) º - s3 + 4.75s2 + 15.84s + 20
Its unit-step response is plotted in Figure 9.9 with the dashed line. Compared with
the quadratic optimal design, the ITAE design has a faster response and a smaller
overshoot. Thus for this problem, the ITAE optimal system is more desirable.
Example 9.6.4
G(s) =
s + 3
---
s(s - 1)
Find an ITAE zero-position-error optimal system. It is also required that the actuating
signal u(t) due to a unit-step reference input meet the constraint lu(t)I :5 10, for
t 2:: O. The pole-zero excess of G(s) is 1 and G(s) has no non-minimim-phase zero;
therefore, the IT AE optimal transfer function
G0(s) =
Wo
S + Wo
(9.57)
is implementable. We find by computer simulation that if w0 = 10, then G0(s) meets
the design specifications. lts step response and actuating signal are plotted in Figure
9 .1 O with the dashed line. They are almost indistinguishable from those of the quad-ratic
optimal system. Because G0(s) <loes not contain plant zero (s + 3), its imple-mentation
will involve the pole-zero cancellation of (s + 3), as will be discussed in
the next chapter. Because it is a stable pole and has a fairly small time constant, its
cancellation will not affect seriously the behavior of the overall system, as will be
demonstrated in the next chapter.
In addition to (9.57), we may also choose the following ITAE optima} transfer
function
Gº(s) =
w6
s
2 + l.4w0s + w
2
(9.58)
lt has pole-zero excess larger than that of G(s) and is implementable. We find by
computer simulation that if Wo = 24.5, then the G0(s) in (9.58) or
600.25
Gº(s) = s2 + 34.3s + 600.25
(9.59)
meets the design specifications. lts step response and actuating signal are plotted in
Figure 9.10 with the dotted lines. The step response is much faster than the ones of
(9.57) and the quadratic optima] system. However, it has an overshoot of about 4.6%.

375 CHAPTER 9 THE INWARD APPROA CH -CHOIC9E. 7 OSEFlE OCVTEIORNA L L B TARSAENOS FOENR EFNUGNICNETIEORINNSG JUOGMENT 375
Example 9.6.5
s 1
G(s) = s(s 2)
Find an ITAE zero-position-error optimal system. lt is also required that the actuating
signal u(t) due to a unit-step reference input meet the constraint /u(t)I :::; 10, for
t 2'.: O. This plant transfer function has a non-minimum-phase zero and no ITAE
standard form is available to carry out the design. However, we can employ the idea
in [34] and use computer simulation to find its ITAE optimal transfer function as [54]
- lO(s - 1)
s2 + 5.ls + 10
(9.60)
under the constraint Ju(t)J :::; 10. We mention that the non-rninimum-phase zero
(s - 1) of G(s) must be retained in G0(s), otherwise G0(s) is not implementable. lts
step response is plotted in Figure 9.11 with the dashed line. lt has a faster response
than the one of the quadratic optimal system in (9.48); however, it has a larger
undershoot and a larger overshoot. Therefore it is difficult to say which system is
better.
9.7 SELECTION BASED ON ENGINEERING JUDGMENT
In the preceding sections, we introduced two criteria for choosing overall transfer
functions. The first criterion is the minimization of the quadratic performance index.
The main reason for choosing this criterion is that it renders a simple and straight-forward
procedure to compute the overall transfer function. The second criterion is
the minimization of the integral of time multiplied by absolute error (ITAE). It was
chosen in [33] because it has the best selectivity. This criterion, however, does not
render an analytical method to find the overall transfer function; it is obtained by
trial and error and by computer simulation. In this section, we forego the concept of
minimizat ion or optimization and select overall transfer functions based on engi-neering
judgment. We require the system to have a zero position error anda good
transient performance. By a good transient performance, we mean that the rise and
settling times are small and the overshoot is also small. Without comparisons, it is
not possible to say what is small. Fortunately, we have quadratic and ITAE optimal
systems for comparisons. Therefore, we shall try to find an overall system that has
a comparable or better transient performance than the quadratic or ITAE optimal
system. Whether the transient performance is comparable or better is based on en-gineering
judgment; no mathematical criterion will be used. Consequently, the se-lection
will be subjective and the procedure of selection is purely trial and error.

376 CHAPTER 9 THE INWARD APPROA CH -CHOIC9E. 7 OSEFlE OCVTEIORNA L L B TARSAENOS FOENR EFNUGNICNETIEORINNSG JUOGMENT 375
Example 9.7.1
Consider the plant transfer function in (9.39), or
2
G(s) =
s(s2 + 0.25s + 6.25)
Wé use computer simulation to select the following two overall transfer functions
G (s) -
20
20
(9.61) 01
- (s + 2)(s2 + 2.5s + 10) s3 + 4.5s2 + l5s + 20
Go2(s) =
20 20
(9.62)
(s + 10)(s2 + 2s + 2) s3 + 12s2 + 22s + 20
The actuating signals of both systems due to a unit-step reference input meet the
constraint luCt)I :s 10 for t 2: O. Their step responses are plotted in Figure 9.9 with,
respectively, the dotted line and the dashed-and-dotted line. The step response of
G01(s) lies somewhere between those of the quadratic optima} system and the ITAE
optimal system. Therefore, G01(s) is a viable altemative of the quadratic or ITAE
optimal system.
The concept of dominant poles can also be used to select G0(s). Consider the
overall transfer function in G02(s). It has a pair of complex-conjugate poles at
- 1 ± j 1 and a real pole at - 10. Because the response due to the real pole dies
out much faster than does the response due to the complex-conjugate poles, the
response of G02(s) is essentially determined or dominated by the complex-conjugate
poles. The complex-conjugate poles have the damping ratio 0.707, and consequently
the step response has an overshoot of about 5% (see Section 7.2.1), as can be seen
from Figure 9.9. However, because the product of the three poles must equal 20 in
order to meet the constraint on the actuating signal, if we choose the nondominant
pole far away from the imaginary axis, then the complex-conjugate poles cannot be
too far away from the origin ofthe s-plane. Consequently, the time constant of G02(s)
is larger than that of G01 (s) and its step response is slower, as is shown in Figure
9.9. Therefore for this problem, the use of dominant poles does not yield a satisfac-tory
system. We mention that if the complex-conjugate poles are chosen at
- 2 ± j2, then the system will respond faster. However, the real pole must be chosen
as 20/8 = 2.5 in order to meet the constraint on the actuating signal. In this case,
the complex-conjugate poles no longer dominate over the real pole, and the concept
of dominant poles cannot be used.

377 CHAPTER 9 THE INWARD APPROA CH -CHO9.I C7 E S E LOEFC TOIVOENR ABLALS E TDR A ONNS FEENRG IFNUENECRTIINOGNS J UDGMENT 377
Example 9.7.2
Consider the plant transfer function in (9.44), or
G(s) =
s + 3
---
s(s - 1)
(9.63)
We have designed a quadratic optimal system in (9.46) and two ITAE optima! sys-tems
in (9.57) with Wo = 10 and (9.59). Their step responses are shown in Figure
9.10. Now using computer simulation, we find that the following
G (s) _ 784
º - s2 + 50.4s + 784
(9.64)
has the response shown with the dashed-and-dotted Iine in Figure 9.10. It is com-parable
with that of the ITAE optima! system in (9.59) under the same constraint on
the actuating signa!. Thus, the overall transfer function in (9.64) can also be used,
although it is not optima! in any sense.
Example9.7.3
s - 1
G(s) = ---
s(s - 2)
We have designed a quadratic optimal system in (9.48) andan ITAE optima! system
in (9.60). Their step responses are shown in Figure 9.11. Now we find, by using
computer simulation, that the response of
- lO(s - 1)
Gº(s) = -(s_+_Vlo _ 1 _ 0 _ )2 (9.65)
lies somewhere between those of (9.48) and (9.60) under the same constraint on the
actuating signal. Therefore, (9.65) can also be chosen asan overall transfer function.
In this section, we have shown by examples that it is possible to use computer
simulation to select an overall transfer function whose performance is comparable
to that of the quadratic or ITAE optima! system. The method, however, is a trial-and-
error method. In the search, we vary the coefficients of the quadratic or ITAE
system and see whether or not the performance could be improved. If we do not
have the quadratic or ITAE optima! system as a starting point, it would be difficult
to find a good system. Therefore, the computer simulation method cannot replace
the quadratic design method, nor the standard forms of the ITAE optimal systems.
It can be used to complement the two optima! methods.

378 CHAPTER 9 THE INWARD APPROA CH -CHO9.I C7 E S E LOEFC TOIVOENR ABLALS E TDR A ONNS FEENRG IFNUENECRTIINOGNS J UDGMENT 377
9 .8 SUMMARYANO CONCLUDING REMARKS
This chapter introduced the inward approach to design control systems. In this ap-proach,
we first find an overall transfer function to meet design specifications and
then implement it. In this chapter, we discussed only the problem of choosing an
overall transfer function. The implementation problem is discussed in the next
chapter.
The choice of an overall transfer function is not entirely arbitrary; otherwise we
may simply choose the overall transfer function as 1. Given a plant transfer function
G(s) = N(s)/D(s), an overall transfer function G0(s) = N0(s)/D0(s) is said to be
implementable if there exists a configuration with no plant leakage such that G0(s)
can be built using only proper compensators. Furthermore, the resulting system is
required to be well posed and totally stable-that is, the closed-loop transfer function
of every possible input-output pair of the system is proper and stable. The necessary
and sufficient conditions for G0(s) to be implementable are that (1) G0(s) is stable,
(2) G0(s) contains the non-minimum-phase zeros of G(s), and (3) the pole-zero ex-cess
of G is equal to or larger than that of G(s). These constraints are not stringe nt ;
0(s)
poles of G0(s) can be arbitrarily assigned so long as they all lie in the open left half
s-plane; other than retaining all zeros outside the region C in Figures 6.13 or 7.4,
all other zeros of G0(s) can be arbitrarily assigned in the entire s-plane.
In this chapter, we discussed how to choose an implementable overall system
to minimize the quadratic and ITAE performance indices. In using these performance
indices, a constraint on the actuating signa! or on the bandwidth of resulting systems
must be imposed; otherwise, it is possible to design an overall system to have a
performance index as small as desirable and the corresponding actuating signal will
approach infinity. The procedure of finding quadratic optimal systems is simple and
straightforward; after computing a spectral factorization, the optimal system can be
readily obtained from (9.19). Spectral factorizations can be carried out by iteration
without computing any roots, or computing ali the roots of (9.16) and then grouping
the open left half s-plane roots. ITAE optimal systems are obtainable from Tables
9.1 through 9.3. Because the tables are not exhaustive, for sorne plant transfer func-tions
(for example, those with non-minimum-phase zeros), no standard forms are
available to find ITAE optimal systems. In this case, we may resort to computer
simulation to find an ITAE optimal system.
In this chapter, we also showed by examples that overall transfer functions that
have comparable performance as quadratic or ITAE optima! systems can be obtained
by computer simulation without minimizing any mathematical performance index.
It is therefore suggested that after obtaining quadratic or ITAE optimal systems, we
may change the parameters of the optima! systems to see whether a more desirable
system can be obtained. In conclusion, we should make full use of computers to
carry out the design.
We give sorne remarks conceming the quadratic optima! design to conclude this
chapter.
l. The quadratic optimal system in (9.19) is reduced from a general formula in
Reference [10]. The requirement of implementability is included in (9.19). lf no

9.8 SUMMARY AND CONCLUDING REMARKS 379
such requirement is included, the optima ! transfe r function that minimizes (9.15)
with r(t) = 1 is
(9.66)
where N + (s) is N(s) with all its right-half-plane roots reftected into the left half
plane. In this case, the resulting overall transfer function may not be imple-mentable.
For example, if G(s) = (s - l)/s(s + 1), then the optima! system
that minimizes
J = Loo [q(y(t) - 1)2 + u2(t)]dt (9.67)
with q = 9 is
3(s + 1)
s2 + 4s + 3
which does not retain the non-minimum-phase zero and is not implementable.
For this optima! system, J can be computed as J = 3. See Chapter 11 of Ref-erence
[12] for a discussion of computing J. The implementable optimal system
that minimizes J in (9.67) is
-3(s - 1)
s2 + 4s + 3
For this implementa~e G0(s), J can b~ computed as J = 21. lt is considerably
larger than the J for G0(s). Although G0(s) has a smaller performance index, it
cannot be implemented.
2. If r(t) in (9.15) is a ramp function, that is r(t) = at, t ~ O, then the optima !
system that minimizes (9.15) is
(9.68)
where
and d [ N(-s) J
k2 = ds Dº(-s) s=O
and
The optima! system in (9.68) is not implementable because it violates the pole-zero
excess inequality. However, if we modify (9.68) as
- ( k2 ) qN(O) N(s)
Go(s) = l + k/ D0(0) D0(s) + ESn+I
(9.69)

•
380 CHAPTER 9 THE INWARD APPROACH --CHOICE OF OVERALL TRANSFER FUNCTIONS
where n : = deg D0(s) and E is a very small positive number, then G0(s) will be
implementable. Furthermore, for a sufficiently small E, Do(s) + Esn +
1 is Hur-witz,
and the frequency response of G0(s) is very close to that of G0(s) in (9.68).
Thus (9.69) is a simple and reasonable modification of (9.68).7
3. The quadratic optimal design can be carried out using transfer functions or using
state-variable equations. In using state-variable equations, the concepts of con-trollability
and observability are needed. The optimal design requires solving an
algebraic Riccati equation and designing a state estimator (see Chapter 11). For
the single-variable systems studied in this text, the transfer function approach
is simpler and intuitively more transparent. The state-variable approach, how-ever,
can be more easily extended to multivariable systems.
PROBLEMS
9.1. Given G(s) = (s + 2)/(s - 1), is G0(s) = 1 implementable? Given G(s) =
(s - l)/(s + 2), is G0(s) = 1 implementable?
9.2. Given G(s) = (s + 3)(s - 2)/ s(s + 2)(s - 3), which of the following G0(s)
are implementable ?
s - 2
s + 3 s - 2
s(s + 2) (s + 2)(s - 3) (s + 2)2
(s + 4)(s - 2) s 2
s4 + 4s2 + 3s + 6 s3 + 4s + 2
9.3. Considera plant with transfer function G(s) = (s + 3)/ s(s - 2).
a. Find an implementable overall transfer function that has all poles at - 2
and has a zero position error.
b. Find an implementable overall transfer function that has ali poles at - 2
and has a zero velocity error. Is the choice unique? Do you have to retain
s + 3 in G0(s)? Find two sets of solutions: One retains s + 3 and the other
does not.
9.4. Considera plant with transfer function G(s) = (s - 3)/ s(s - 2).
a. Find an implementable overall transfer function that has all poles at - 2
and has a zero position error.
b. Find an implementable overall transfer function that has ali poles at - 2
and has a zero velocity error. Is the choice unique if we require the degree
of G0(s) to be as small as possible?
7This
modification was suggested by Professor Jong-Lick Lin of Cheng Kung University, Taiwa n.

PROBLEMS 381
9.5. What types of reference signals will the following GJs) track without an error?
2
•
-5s - 2
a. G0(s) = --2-----
- s - 5s - 2
4s2 + s + 3
b. G0(s) =
s5 + 3s4 + 4s + s + 3
- 2s2 + 154s + 120
e G (s) = ------------
0 s4 + 14s3 + 7ls2 + 154s + 120
9.6. Consider two systems. One has a settling time of 10 seconds and an overshoot
of 5%, the other has a settling time of 7 seconds andan overshoot of 10%. Is
it possible to state which system is better? Now we introduce a performance
indexas
J = k, · (Settling time) + k2 · (Percentage overshoot)
lf k, = k2 = 0.5, which system is better? If k1 = 0.8 and k2 = 0.2, which
system is better?
9.7. Is the function
J = L''' [q(y(t) - r(t)) + u(t)]dt
with q > O a good performance criterion?
9.8. Consider the design problem in Problem 7.15 ora plant with transfer function
G(s) = - 2/ s2. Design an overall system to minimize the quadratic per-
- formance index in (9.15) with q = 4. What are its position error and velocity
error?
9.9. In Problem 9.8, design a quadratic optimal system that is as fast as possible
under the constraint that the actuating signal due to a step-reference input must
have a magnitude less than 5.
9.10. Plot the poles of G0(s) as a function of q in Problem 9.9.
9.11. Consider the design problem in Problem 7 .14 or a plant with transfer function
0.015
G(s) =s _2_+_0 1 _1s_+_ o3
Design an overall system to minimize the quadratic performance index in
(9.15) with q = 9. Is the position error of the optimal system zero? Is the
index of the optimal system finite?
9.12. Consider the design problem in Problem 7.12 ora plant with transfer function
4(s + 0.05)
G(s) =
s(s + 2)(s - 1.2)
Find a quadratic optima} system with q = 100. Carry out the spectral factor-ization
by using the iterative method discussed in Section 9.4.2.

382 CHAPTER 9 THE INWARD APPROA CH -CHOICE OF OVERALL TRANSFER FUNCTIONS
9.13. Let Q(s) = D0(s)D0( -s) with
Q(s) = ªo + a2s2 + a4s4 + ... + ª2ns2n
and
Show
ªo b'6
a2 2b0b2 - hi
a4 2b0b4 - 2b1b3 + b~
a2n = 2b0b2n - 2b1b2n- l + 2b2h2n-2 - · · · + (- ltb~
where b, = O, for i > n.
9.14. The depth of a submarine can be maintained automatically by a control system,
as discussed in Problem 7.8. The transfer function of the submarine from the
stem angle () to the actual depth y can be approximated as
G(s) - lO(s + 2)2
- (s + 10)(s2 + 0.1)
Find an overall system to minimize the performance index
1 = fo"' [(y(t) - 1)2 + e2Jdt
9.15. Considera plant with transfer function s/(s2 - 1). Designan overall system
to minimize the quadratic performance index in (9.15) with q = l. Does the
optimal system have zero position error? If not, modify the overall system to
yield a zero positio n error.
9.16. Consider a plant with transfer function G(s) = 1/ s(s + 1). Find an imple-mentable
transfer function to minimize the ITAE criterion and to have zero
position error. It is also required that the actuating signal due to a unit-step
reference input have a magnitude less than 10.
9.17. Repeat Problem 9.16 with the exception that the overall system is required to
have a zero velocity error.
9.18. RepeatProblem9.16forG(s) = 1/s(s - 1).
9.19. RepeatProblem9.17forG(s) = l/s(s - 1).
9.20. Find an ITAE zero-position-error optimal system for the plant given in Problem
9.8. The magnitude of the actuating signa! is required to be no larger than the
one in Problem 9.8.

PROBLEMS 383
9.21. Find an ITAE zero-position-erroroptimal system for the plant in Problem 9.11.
The real part of the pales of the optimal system is required to equal that in
Problem 9.11.
9.22. Is it possible to obtain an ITAE optima! system for the plant in Problem 9.12
from Table 9.1 or 9.2? If yes, what will happen to the plant zero?
9.23. Repeat Problem 9.22 for the plant in Problem 9.14.
9.24. o. Considera plant with transfer function G(s) = (s + 4)/ s(s + 1). Design
an ITAE zero-position-error optimal system of degree l. It is required that
the actuating signal due to a unit-step reference input have a magnitude less
than 10.
b. Considera plant with transfer function G(s) = (s + 4)/s(s + 1). Design
an ITAE zero-position-error optima! system of degree 2. It is required that
the actuating signal dueto a unit-step reference input have a magnitude less
than 10.
c. Compare their unit-step responses.
9.25. Consider the generator-motor set in Figure 6.1. lts transfer function is assumed
to be
G(s) - ---------
300
- s4 + 184s3 + 760s2 + 162s
It is a type 1 transfer function. Design a quadratic optima! system with q
25. Designan ITAE optima! system with u(O+) = 5. Plot their poles. Are
there many differences?
9.26. Considera plant with transfer function 1/ s2• Find an optima! system with zero
velocity error to minimize the ITAE criterion under the constraint lu(t)I ::::: 6.
[Answer: (6s + 2.5)/(s3 + 2.38s2 + 6s + 2.5).]
9.27. If software for computing step responses is available, adjust the coefficients of
the quadratic optima! system in Problem 9.8, 9.11, 9.12, 9.14, or 9.15 to see
whether a comparable or better transient performance can be obtained.

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